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PHP MySQL display image from database

Hi I have form to make new article with informations and image. I susccesfuly save all info and image to database (i guess). And when I want to display info and image so only info works. See in picture.

Any help? Thanks a lot

图片

Inserting_post.php

<!DOCTYPE html>
    <html lang="en">
 <head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>APK Market</title>
<link rel="stylesheet" type="text/css" href="../style/style.css">
<link href="../bootstrap/css/bootstrap.css" rel="stylesheet" media="screen">
<script src="../bootstrap/js/bootstrap.min.js"></script>
<script src="../bootstrap/js/bootstrap.js"></script>
 </head>
 <body>

 <form method="post" action="insert_post.php" enctype="multipart/form-data">

<div class="new_post">
<div class="headtitle">Insert new post</div>
 <div class="new_title">
<p>New title</p>
<input type="text" name="title">
 </div>
 <div class="new_author">
<p>Author</p>
<input type="text" name="author">
 </div>
 <div class="new_keywords">
<p>Keywords</p>
<input type="text" name="keywords">
   </div>
   <div class="new_image">
<p>Image</p>
<input type="file" name="image">
  </div>
  <div class="new_content">
<textarea name="content" cols="20" rows="8"></textarea>
   </div>
  <div class="submit">
<input type="submit" name="submit" value="OK">
  </div>
  </div>
 </form>
  <script src="https://code.jquery.com/jquery.js"></script>
  <script src="../bootstrap/js/bootstrap.js"></script>
  </body>
  </html>
   <?php
   include("../includes/connect.php");

   if(isset($_POST['submit']))
   {
$games_date = date('y-m-d-h');
$games_title = $_POST['title'];
$games_author = $_POST['author'];
$games_keywords = $_POST['keywords'];
$games_image = $_FILES['image']['name'];
$games_tmp = $_FILES['image']['tmp_name'];
$games_content = $_POST['content'];

if($games_title=="" or $games_author=="" or $games_keywords==""
     or $games_content=="")
{
    echo"<script>alert('any field is empty')</script>";
    exit();
}
else 
move_uploaded_file($games_tmp,"../uploaded_images/$games_image");
$insert_query= "insert into games(games_title,games_date,games_author,games_image,
    games_keywords,games_content)

      values ('$games_title','$games_date','$games_author','$games_image',
   '$games_keywords','$games_cont         ent')";

      }
         if(mysql_query($insert_query))
{
    echo "<center><h1>Post published seccesfuly!</h1></center>";
}
     ?>

     display page:

      <div class="content">
   <?php
     include('connect.php');

    $select_posts = "select * from games";
   $run_posts = mysql_query($select_posts);

   while($row=mysql_fetch_array($run_posts))
     {
echo '<p class="games_title_result">' .$games_title = $row['games_title'];
echo '<p class="games_image_result"><img src="<?php echo $row["games_image"];?>';
echo '<p class="games_content_result">' .$games_content = $row['games_content'];
echo '<p class="games_date_result">' .$games_date = $row['games_date'];
echo '<p class="games_author_result">' .$games_author = $row['games_author'];
     }
    ?>  
    </div>

Here's my answer:

echo '<p class="games_image_result"><img src="uploaded_images/'.$row["games_image"].'" />';

You got a syntax error where you include a <?php and ?> inside your php, and on your echo at that.

So when you look at your img's src, it would have those.

Also, you forgot to close your img tag and your other p tags.

Your image tag isn't closed properly and the path to the image is not complete. You are only specifying the filename. I am not sure about your directory structure

echo '<p class="games_image_result"><img src="uploaded_images/<?php echo $row["games_image"];?>" />';

Display Image from DataBase using php

           # SQL Statement
           $sql = "SELECT `idimage` FROM `yourTableName` WHERE id=" . mysqli_real_escape_string($_GET['id']) . ";";
           $result = mysqli_query("$sql") or die("Invalid query: " . mysqli_error());

           # Set header
           header("Content-type: image/your image name which type of your image");
           echo mysqli_result($result, 0);
      }
      else
           echo 'Please check the ID!';
 ?>

I wish it's may be help you

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